Problem: Boolean Expression Compressor

You are asked to build a compressor for Boolean expressions that transforms expressions to the shortest form keeping their meaning.

The grammar of the Boolean expressions has terminals `0` `1` `a` `b` `c` `d` `-` `^` `*` `(` `)`, start symbol <E> and the following production rule:

> <E>  ::=  `0`  |  `1`  |  `a`  |  `b`  |  `c`  |  `d`  |  `-`<E>  |  `(`<E>`^`<E>`)`  |  `(`<E>`*`<E>`)`

Letters `a`, `b`, `c` and `d` represent Boolean variables that have values of either `0` or `1`. Operators are evaluated as shown in the Table below. In other words, `-` means negation (NOT), `^` means exclusive disjunction (XOR), and `*` means logical conjunction (AND).

Table: Evaluations of operators
<image>

Write a program that calculates the length of the shortest expression that evaluates equal to the given expression with whatever values of the four variables.

For example, `0`, that is the first expression in the sample input, cannot be shortened further. Therefore the shortest length for this expression is 1.

For another example, `(a*(1*b))`, the second in the sample input, always evaluates equal to `(a*b)` and `(b*a)`, which are the shortest. The output for this expression, thus, should be `5`.

Input

The input consists of multiple datasets. A dataset consists of one line, containing an expression conforming to the grammar described above. The length of the expression is less than or equal to 16 characters.

The end of the input is indicated by a line containing one `.` (period). The number of datasets in the input is at most 200.

Output

For each dataset, output a single line containing an integer which is the length of the shortest expression that has the same value as the given expression for all combinations of values in the variables.

Sample Input


0
(a*(1*b))
(1^a)
(-(-a*-b)*a)
(a^(b^(c^d)))
.


Output for the Sample Input


1
5
2
1
13






Example

Input

0
(a*(1*b))
(1^a)
(-(-a*-b)*a)
(a^(b^(c^d)))
.


Output

1
5
2
1
13

Solution (in CPP):

Solution:
#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = (1e+18) + 7;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-6;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

ll mod_pow(ll a, ll n) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%mod;
		a = a * a%mod; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

//ll inv(ll a, ll p) {
//	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
//}
//modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

int dx[4] = { 1,0,-1,0 };
int dy[4] = { 0,1,0,-1 };

bool isnum(char &t) {
	return '0' <= t && t <= '9';
}
bool isalf(char &t) {
	return 'a' <= t && t <= 'z';
}

int b[4];
int expr(string &s, int &i);

int expr(string &s, int &i) {
	if (isnum(s[i])) {
		int res = s[i] - '0'; i++;
		return res;
	}
	if (isalf(s[i])) {
		int res = b[s[i] - 'a']; i++;
		return res;
	}
	if (s[i] == '-') {
		i++;
		return !expr(s, i);
	}
	if (s[i] == '(') {
		i++;
		int le = expr(s, i);
		char t = s[i];
		i++;
		int ri = expr(s, i);
		i++;
		if (t == '*')return le & ri;
		else return le ^ ri;
	}
}


vector<P> dp[4];
const int mn = (1 << 16);

int ans[mn];

void init() {
	dp[0].push_back({ 0,1 });
	dp[0].push_back({ (1 << 16) - 1,1 });
	rep(i, 4) {
		int s = 0;
		rep(j, 16) {
			if (j&(1 << i)) {
				s += (1 << j);
			}
		}
		dp[0].push_back({ s,1 });
		dp[0].push_back({ mn - 1 - s,2 });
	}
	rep1(i, 3) {
		rep(j, i) {
			int k = i - 1 - j;
			if (j > k)continue;
			rep(x, dp[j].size()) {
				rep(y, dp[k].size()) {
					int ls = dp[j][x].first; int lc = dp[j][x].second;
					int rs = dp[k][y].first; int rc = dp[k][y].second;
					dp[i].push_back({ ls&rs,lc + rc+3 });
					dp[i].push_back({ mn - 1 - (ls&rs),lc + rc + 4 });
					dp[i].push_back({ ls^rs,lc + rc+3 });
					dp[i].push_back({ mn - 1 - (ls^rs),lc + rc+4 });
				}
			}
		}
	}
	fill(ans, ans + mn, mod);
	rep(i, 4)for (P p : dp[i]) {
		ans[p.first] = min(ans[p.first], p.second);
	}
}

void solve() {
	string in;
	while (cin >> in) {
		if (in == ".")break;
		int s = 0;
		rep(i, 16) {
			rep(j, 4) {
				if (i&(1 << j))b[j] = true;
				else b[j] = false;
			}
			int k = 0;
			int val = expr(in, k);
			assert(k == in.size());
			if (val)s += (1 << i);
		}
		cout << ans[s] << endl;
	}
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(7);
	init();
	//int t; cin >> t; rep(i, t)solve();
	solve();
	stop
		return 0;
}